We will first consider the case in which the line segments may intersect. Let's define a "region" as a contiguous sequence of equal ~L~ values (making sure to count ~L_N~ and ~L_1~ as being consecutive). Let ~r~ be the number of regions (which can be easily counted in ~\mathcal O(N)~ time), and note that ~r~ is guaranteed to be at least ~2~. There are then ~r~ points on the circle which are immediately surrounded by two different ~L~ values, meaning that at least one line segment must pass through the circle there. This suggests that there must be at least ~r~ line segment endpoints, and so at least ~c = \lceil \frac r 2 \rceil~ line segments. Furthermore, ~c~ line segments are always enough to get the job done. Imagine sorting the ~r~ points mentioned above, adding one more arbitrary point if ~r~ is odd, and then connecting the first point to the ~c~-th one, the second one to the ~(c+1)~-st one, and so on. This will divide up the circle such that each region is segmented off entirely from the others.

The non-intersecting case isn't so simple, but can be solved with dynamic programming. Let ~DP[i][s]~ be the minimum number of line segments required to divide up the portion of the circle starting at index ~i~ and spanning ~s~ markings, with ~L_i~ being the portion's "representative marking". We'd like to consider various ways of cutting this portion into smaller sub-portions while saving on cuts when those sub-portions have markings matching ~L_i~. Let's consider all possible splits of this subarray into two subarrays, with the first including the first ~k~ markings and the second including the remaining ~s-k~ markings ~(1 \le k < s)~. Note that the second subarray starts at marking ~i+k~ (wrapping around if it exceeds ~N~).

If ~k > 1~, we'll make one cut to separate index ~i~ from the remainder of the first subarray, which will then require ~DP[i+1][k-1]~ more cuts itself. Similarly, we'll make one more cut to separate out the second subarray, which will then require ~DP[i+k][s-k]~ more cuts itself. However, each of these two main cuts can be omitted if that portion's representative marking (~L_{i+1}~ or ~L_{i+k}~) is equal to ~L_i~. We can calculate ~DP[i][s]~ by determining the above value for all possible values of ~k~ and taking the minimum.

Finally, any arbitrary marking can be used as the "representative marking" of the entire circle, meaning that the answer is for example ~DP[1][N]~. The approach described above involves ~\mathcal O(N^2)~ DP states and ~\mathcal O(N)~ transitions per state, resulting in a time complexity of ~\mathcal O(N^3)~.